Hernalt

2011-May-30, 11:28 AM

:wall:

The blue chart on Wikipedia Delta-V (http://en.wikipedia.org/wiki/Delta-v) shows 2.5 km/s LEO to GTO and 0.7 GTO to Earth C3.

In trying to reproduce these two numbers, this is what I get:

v_{GTO Injection} - v_{LEO} = \sqrt{\mu_{earth}\(\frac{2}{r_{LEO}}-\frac{1}{a_{GTO}})}-\sqrt{\mu_{earth}\frac{1}{r_{LEO}}}

= 10.24 km/s - 7.78 km/s = 2.45 km/s

v_{GTOExitC3} - v_{GTOExit} = \sqrt{2}\sqrt{\mu_{earth}\(\frac{2}{r_{GEO}}-\frac{1}{a_{GTO}})}-\sqrt{\mu_{earth}\(\frac{2}{r_{GEO}}-\frac{1}{a_{GTO}})}

= 2.26 km/s - 1.60 km/s = .66 km/s

Am I doing that correctly? The progression implies you don't actually circularize orbit "into" GEO. I interpret that you just keep on trucking up "past" GEO.

(And / also / or) How do you do Earth C3 to Mars Transfer? Does Earth C3 by convention represent a particular orbital radius? It seems like it's suitable for any r using \sqrt{2} r. Coming out of a Hohmann transfer is - as far as I can tell for purposes of calculating escape velocity - exactly tangent to the circular orbit.

I did a calculation with \mu_{sun}, r_{earth orbit}, v_{earth orbit}, r_{mars orbit}, v_{mars orbit}, a_{earth mars hohmann} and got 5.5 km/s, which is much greater than the Wiki page's 0.6 km/s + 0.9 km/s. So I've got naive assumptions in there. It made sense to me that if you're outside the gravity well of both earth and mars, then neither \mu_{earth} or \mu_{mars} are used.

What obvious n00b fubars am I making? :doh: Thanks.

The blue chart on Wikipedia Delta-V (http://en.wikipedia.org/wiki/Delta-v) shows 2.5 km/s LEO to GTO and 0.7 GTO to Earth C3.

In trying to reproduce these two numbers, this is what I get:

v_{GTO Injection} - v_{LEO} = \sqrt{\mu_{earth}\(\frac{2}{r_{LEO}}-\frac{1}{a_{GTO}})}-\sqrt{\mu_{earth}\frac{1}{r_{LEO}}}

= 10.24 km/s - 7.78 km/s = 2.45 km/s

v_{GTOExitC3} - v_{GTOExit} = \sqrt{2}\sqrt{\mu_{earth}\(\frac{2}{r_{GEO}}-\frac{1}{a_{GTO}})}-\sqrt{\mu_{earth}\(\frac{2}{r_{GEO}}-\frac{1}{a_{GTO}})}

= 2.26 km/s - 1.60 km/s = .66 km/s

Am I doing that correctly? The progression implies you don't actually circularize orbit "into" GEO. I interpret that you just keep on trucking up "past" GEO.

(And / also / or) How do you do Earth C3 to Mars Transfer? Does Earth C3 by convention represent a particular orbital radius? It seems like it's suitable for any r using \sqrt{2} r. Coming out of a Hohmann transfer is - as far as I can tell for purposes of calculating escape velocity - exactly tangent to the circular orbit.

I did a calculation with \mu_{sun}, r_{earth orbit}, v_{earth orbit}, r_{mars orbit}, v_{mars orbit}, a_{earth mars hohmann} and got 5.5 km/s, which is much greater than the Wiki page's 0.6 km/s + 0.9 km/s. So I've got naive assumptions in there. It made sense to me that if you're outside the gravity well of both earth and mars, then neither \mu_{earth} or \mu_{mars} are used.

What obvious n00b fubars am I making? :doh: Thanks.